HOW TO EVALUATE INTEGRALS USING SUBSTITUTION

Question 1 :

Integrate the following with respect to x

cot x/log (sin x)

Solution :

=  ∫cot x/log (sin x) dx 

Let u = log (sin x)

Differentiating with respect to "x"

du/dx  =  1/sin x ( cos x)

du  =  cos x / sin x

du  =  cot x dx

∫cot x/log (sin x) dx  =  ∫du/u

=  log u + c

=  log (log (sin x)) + c

Question 2 :

Integrate the following with respect to x

cosec x/log (tan (x/2))

Solution :

=  ∫ [cosec x/log (tan (x/2))] dx 

Let u  =  log (tan x/2)

du  =  (1/tan (x/2)) ⋅ sec² (x/2) (1/2) dx

  =  (cos x/2 / sin x/2) ⋅ (1/cos² (x/2)) (1/2) dx

  =  (1 / 2 sin x/2 cos (x/2))dx

  =  1/ sin x dx 

du  =   cosec x dx

∫ [cosec x/log (tan (x/2))] dx   =  ∫ du/u

  =  log u + c

  =  log (log tan x/2) + c

Question 3 :

Integrate the following with respect to x

sin 2x / (a² + b² sin²x)

Solution :

=  ∫ [sin 2x / (a² + b² sin²x)] dx 

Let u  =  a² + b² sin²x

  du  =  0 + b² (2 sin x cos x) dx

  du  =   b² sin 2x dx

 sin 2x dx  =  du/b²

∫ [sin 2x / (a² + b² sin²x)] dx  =  ∫ (du/b²)/u

  =  (1/b²) ∫ (1/u) du

  =  (1/b²) log u + c

  =  (1/b²) log (a² + b² sin²x) + c

Question 4 :

Integrate the following with respect to x

sin^-1x/√(1 - x²) 

Solution :

=  ∫ [sin^-1x/√(1 - x²)] dx 

Let u = sin^-1x

du  =  (1/√(1 - x²)) dx

∫ [sin^-1x/√(1 - x²)] dx  =  ∫ u du

  =  u²/2 + c

  =  (sin^-1x)²/2 + c

Question 5 :

Integrate the following with respect to x

√x/(1 + √x)

Solution :

=  ∫ [√x/(1 + √x)] dx 

Let u = 1 + √x

du  =  (0 + 1/2√x) dx

2√x du  =  dx

∫ [√x/(1 + √x)] dx  =  ∫ [ (u - 1)/u] 2 √x du

  =  ∫ [ (u - 1)/u] 2 (u - 1) du

  =  2 ∫ [ (u - 1)²/u] du

  =  2 ∫ [u² - 2u + 1)/u] du

  =  2 [(∫u du) - 2 ∫ du + ∫(1/u) du]

=  2 [u²/2 - 2u + log u] + c

=  u² - 4u + 2log u + c

=  (1 + √x)² - 4(1 + √x) + 2log (1 + √x) + c

Question 6 :

Integrate the following with respect to x

1/(x log x log (log x))

Solution :

=  ∫ [1/(x log x log (log x))] dx 

Let u  =  log (log x)

du  =  (1/log x) (1/x)

du  =  (1/x log x) dx

∫ [1/(x log x log (log x))] dx = ∫ 1/(x logx) (log (logx))] dx 

  =  ∫ (du/u)

  =  log u 

  =  log (log (log x)) + c

Question 7 :

Integrate the following with respect to x

1/ √(8x - x²) dx

Solution :

= 1/ √(8x - x²) dx

8x - x² = - x² + 8x

= -(x² - 2x(4))

= -[x² - 2x(4) + 4² - 4²]

= -[(x - 4)² - 4²]

(8x - x²) = [4² - (x - 4)²]

 √(8x - x²) = √[4² - (x - 4)²]

Let u = (x - 4)

du = dx

∫1/ √(8x - x²) dx = ∫ 1/ √[4² - u²] du

= sin^-1(u/4) + C

Applying the value of u, we get

= sin^-1[(x - 4)/4) + C

Question 8 :

Integrate the following with respect to x

(x³ + x²)(3x² + 2x)

Solution :

∫ (x³ + x²)(3x² + 2x) dx

Let u = x³ + x²

du = 3x² + 2x

∫ (x³ + x²)(3x² + 2x) dx = ∫ u du

= u²/2 + C

Applying the value of u, we get

= (x³ + x²)²/2 + C

Question 9 :

Integrate the following with respect to x

1/(cos²x)√(1 + tan x) dx

Solution :

∫ 1/(cos²x)√(1 + tan x) dx

Let u = 1 + tan x

du = 0 + sec² x dx

du = sec² x dx

du = 1/cos² x dx

∫ 1/(cos²x)√(1 + tan x) dx = ∫  √u du

= u^3/2 / (3/2) + C

= (2/3) u^3/2 + C

Applying the value of u, we get

= (2/3) (1 + tan x)^3/2 + C

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